sum-of-even-indices.md

Sum of even indices

Problem Description

You are given an array A of length N and Q queries given by the 2D array B of size Q*2. Each query consists of two integers B[i][0] and B[i][1]. For every query, the task is to calculate the sum of all even indices in the range A[B[i][0]…B[i][1]].

Note : Use 0-based indexing

Problem Constraints

1 <= N <= 105
1 <= Q <= 105
1 <= A[i] <= 100
0 <= B[i][0] <= B[i][1] < N

Input Format

First argument A is an array of integers.
Second argument B is a 2D array of integers.

Output Format

Return an array of integers.

Example Input

Input 1:
A = [1, 2, 3, 4, 5]
B = [   [0,2] 
        [1,4]   ]

Input 2:
A = [2, 1, 8, 3, 9]
B = [   [0,3] 
        [2,4]   ]

Example Output

Output 1:
[4, 8]

Output 2:
[10, 17]

Example Explanation

Explanation 1:
The subarray for the first query is [1, 2, 3] whose sum of even indices is 4.
The subarray for the second query is [2, 3, 4, 5] whose sum of even indices is 8.

Explanation 2:
The subarray for the first query is [2, 1, 8, 3] whose sum of even indices is 10.
The subarray for the second query is [8, 3, 9] whose sum of even indices is 17.

Output

function sumOfEvenIndices(A, B){
        // taking pfsum of even number
        const pf = []
        const ans = []
        pf[0] = A[0];
        for(let i = 1 ; i < A.length ; i++){
            if(i % 2 == 0){
                pf[i] = pf[i - 2] + A[i]
            }else{
                pf[i] = pf[i - 1]
            }
        }

        for(let i = 0 ; i < B.length ; i++){
            if(B[i][0] == 0){
                ans[i] = pf[B[i][1]]
            }else{
                ans[i] = pf[B[i][1]] - pf[B[i][0] - 1]
            }
        }
        return ans
}